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Question Number 33352 by caravan msup abdo. last updated on 15/Apr/18

$$letgiveS\left(x\right)=\sum _{n⩾0}\frac{{(-1)}^n}{\sqrt{x+n}},x>0$$$$1)studythecontnuity,derivsbility,limits$$$$at{0}^{+}and+\mathrm{\infty }$$$$2)wegive{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt=\frac{\sqrt{\pi }}{2}.provethat$$$$\mathrm{\forall }x>0S\left(x\right)=\frac{1}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-tx}}{\sqrt{t}(1+e^{-t})}dt.$$

Commented by math khazana by abdo last updated on 20/Apr/18

$$1)letput{f}_n\left(x\right)=\frac{{(-1)}^n}{\sqrt{x+n}}thesequence\left(f_n\right)arecontinue$$$$ndtheserie\mathrm{\Sigma }{f}_n\left(x\right)isconvergent(ifwetake$$$$\phi \left(t\right)=\frac{1}{\sqrt{x+t}}.\phi isdecreszingso\mathrm{\Sigma }{f}_{n}isalternating)$$$$itssumS\left(x\right)iscontinueforx>0alsothe\left(f_n\right)are$$$$derivablesand{f}_n^{{}^{\prime }}\left(x\right)={(-1)}^n\frac{-1}{2(x+n)\sqrt{x+n}}$$$$\sum ^{}{f}_n^{{}^{\prime }}\left(x\right)isconvergentasaalternatingserieso$$$$Sisderivableand{S}^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^{n+1}}{2(x+n)\sqrt{x+n}}.$$$$2)wehave{\int }_0^{\mathrm{\infty }}\frac{e^{-tx}}{\sqrt{t}(1+e^{-t})}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-tx}}{\sqrt{t}}\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-nt}\right)dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}\frac{e^{-(n+x)t}}{\sqrt{t}}dt(ch.\sqrt{t}=u)$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}\frac{e^{-(n+x)u^2}}{u}2udu$$$$=2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-(n+x)u^2}du$$$${=}_{\sqrt{n+x}u=t}2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{n+x}}$$$$=2.\frac{\sqrt{\pi }}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\sqrt{n+x}}=\sqrt{\pi }S\left(x\right)\Rightarrow$$$$S\left(x\right)=\frac{1}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{e^{-tx}}{\sqrt{t}(1+e^{-t})}dt.$$

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